MEASUREMENTS FROM THE ENGLISH TO THE METRIC SYSTEM AND FROM THE METRIC TO THE ENGLISH SYSTEM; AND SOLVING PROBLEMS USING
RATIO, PROPORTION, AND TRIGONOMETRY
TASK 1. Describe the processes for converting linear measurements from the English to the metric system and from the metric to the English system.
CONDITIONS
Within a self-study environment and given the subcourse text, without assistance.
STANDARDS Within two hours REFERENCES
No supplementary references are needed for this task.
1. Introduction
Lesson 1 provided a review of the whol e number syste m and disc ussed the processes for solving machine shop work problems through addition, subtraction, multiplication, and division of fractions and decimals. It also discussed the conversion of decimals to fractions and fractions to decimals. To provide a compl ete coverage of machi ne shop cal culations, Lesson 2 will discuss the processes for converting linear measurements between the English and metric systems, and for solving machine shop problems involving the use of ratio, proportion, and trigonometry. This task will focus on the conversion of linear measurements from the English to the metric system and from the metric to the English system.
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2. Linear Measure
a. General. Linear measure is the measurement of line distance. In machine shop calculations, it is important to know how to convert linear measurements from the English to the metric system and vice-versa.
Subsequent paragraphs provide an explanation of both these systems and how to convert from one system to another.
b. English System . This system consists, basically, of the inch, foot, yard, and mile. The foot is the basic unit of measure. The inch is a subdivision of the foot, while the yard and the mile are multiples of the foot. Table 2 below depicts the English system.
TABLE 2. ENGLISH SYSTEM LINEAR MEASUREMENTS.
12 inches (in) = 1 foot (ft) 3 feet (ft) = 1 yard (yd) 5280 feet (ft) = 1 mile 1760 yards (yd) = 1 mile
c. Metric System . This system is based on the decimal system, just like the United States dollar (10 cents equals a dime, and 10 dimes equal one dollar). The meter is the basic unit of measurement, as depicted in Table 3 below. As shown in this table, units that are multiples or fractional parts of the meter, such as the millimeter, are designated as such by prefixes to the word meter.
TABLE 3. METRIC SYSTEM LINEAR MEASUREMENTS.
10 millimeters (mm) = 1 centimeter (cm) 10 centimeters (cm) = 1 decimeter (dm) 10 decimeters (dc) = 1 meter (m) 1000 meters (m) = 1 kilometer (km)
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3. Conversion
a. Requirements for Conversion. To convert from one system to another, a knowledge of equivalent values between these two systems is necessary.
Table 4 provides a list of equivalent values between the two systems for linear measurements equal to or grea ter than one in ch. Table 5 on the following page provides the metric equivalents of linear measurements of less than an inch.
TABLE 4. LINEAR MEASUREMENTS EQUIVALENT VALUES.
1 mil = 0.001 in
1 mile = 0.868 nautical mile 1 kilometer = .62 mile
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TABLE 5. DECIMAL AND METRIC EQUIVALENTS OF FRACTIONS OF AN INCH.
b. Converting From the English System to the Metric System . (1) Rule 1. To convert miles to kilometers, multiply by 1.61.
EXAMPLE
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(2) Rule 2. To convert yards to meters, multiply by 0.9144.
EXAMPLE
Reduce 3 yards to meters.
3 x 0.9144 = 2.7432 m
(3) Rule 3. To convert inches to centimeters, multiply by 2.54.
EXAMPLE
Reduce 16 inches to centimeters.
16 x 2.54 = 40.64 cm
(4) Rule 4. To convert inches to millimeters, multiply by 25.4.
EXAMPLE
Reduce 2 feet 8 inches to millimeters.
2 ft = 24 in + 8 in = 32 in 32 in x 25.4 = 812.8 mm
(5) Rule 5. To convert inches to millimeters multiply by 25.4.
EXAMPLE
Reduce 11 inches to millimeters.
11 x 25.4 = 279.4 mm
c. Converting from the Metric System to the English System .
(1) Rule 1. To convert kilometers to miles, multiply by 0.62.
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EXAMPLE
Reduce 60 kilometers to miles.
60 x 0.62 = 37.2 miles
(2) Rule 2. To convert meters to yards multiply by 1.0936.
EXAMPLE
Reduce 3.5 meters to yards.
3.5 x 1.0936 = 3.8276 yd
(3) Rule 3. To convert meters to inches multiply by 39.37.
EXAMPLE
Reduce 1.6 meters to inches.
1.6 x 39.37 = 62.99 in
(4) Rule 4. To convert centimeters to inches multiply by 0.3937.
EXAMPLE
Reduce 76.2 centimeters to inches.
76.2 x 0.3937 = 30 in
(5) Rule 5. To convert millimeters to inches multiply by 0.03937.
EXAMPLE
Reduce 88.9 millimeters to inches.
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d. Metric equivalents of le ss than an in ch ar e expressed in te rms of millimeters. To con vert lin ear measurements of less tha n an inc h to millimeters and millimeter linear measurements to fractions of an inch use Table 5 on page 37 for the more common fractions. Otherwise, follow the procedure stated above and in paragraph 5 (Reduction of a common fraction to a decimal fraction), beginning on page 24, and paragraph 6 (Reduction of a decimal fraction to a common fraction), beginning on page 25.
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LESSON 2 CONVERSION OF LINEAR
MEASUREMENTS FROM THE ENGLISH TO THE METRIC SYSTEM AND FROM THE METRIC TO THE ENGLISH SYSTEM; AND SOLVING PROBLEMS USING
RATIO, PROPORTION, AND TRIGONOMETRY
TASK 2. Describe the proc esses for solv ing problems using ratio and proportion.
CONDITIONS
Within a self-study environment and given the subcourse text, without assistance.
STANDARDS Within two hours REFERENCES
No supplementary references are needed for this task.
1. Introduction
A major part of machine shop work involves the fabrication of such parts as a spool, gear, or pulley for machinery and vehicle powertrains. These type parts must be machined to a predetermined size that will enable their turning at a given number of revolutions per minute (rpm). The machining of these parts requires knowledge of the mathematical processes involved in determining the siz e to which the se items must be mac hined. This task, therefore, is designed to provide the processes for determining the size of these parts through the use of mathematical problems involv ing ratio and proportion. Subsequent paragraphs provide an explanation of the methods for solving ratio and proportion problems.
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2. Ratio and Proportion
a. General. Ratio and proportion are methods for reducing the confusion and minimizing the possibilities of error in working arithmetic problems. A working knowledge of these methods makes it easier to solve many shop problems. The language of ratio and proportion is mostly a sign language.
Letters and symbols are commonly used in place of long numbers and represent unknown quantities and values.
b. Ratio.
(1) Ratio is the relation which one quantity bears to another quantity of the same kind. It is used extensively in shop work. Shop drawings or blueprints are generally drawn to scale. Scale means one figure is used to represent another. Usually a small figure represents a larger figure. For example, on a blueprint 1 inch might represent 1 foot.
(2) The two numbers used in the ratio are called the “terms” of the ratio.
The first number of a ratio is called the antecedent; the second number is called the consequent. The consequent is the divisor. The colon (:) is the sign of ratio and means “is to.” Thus, 3 : 5 reads “3 is to 5.” It is in effect a dividing sig n without the dividing (-) line. Such oth er expressions as “in the same ratio”, “in the same proportion’, or “pro rata”
all have the same meaning.
(3) The ratio of one number to another is really the quotient of the first number divided by the second number.
EXAMPLE
Determine the ratio of the expression 8 : 2.
8 : 2 = 4 Divide 8 by 2. Thus, the ratio or value of 8 to 2 is 4.
(4) The value of a ratio is not changed by either multiplying or dividing both terms by the same number.
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EXAMPLE
Multiply the expression 3 : 2 by 2.
3 : 2 = 6 : 4 Multiplying both terms by 2 renders the expression 6 : 4. Check by dividing 3 by 2 which renders a ratio of 1.5. In the second expression dividing 6 by 4 also renders a ratio of 1.5. second expression dividing 2 by 1 renders a ratio of 2. Thus 8 : 4 is equal in value to 2 : 1.
c. Proportion.
(1) Proportion is a statement of equality between two ratios. Thus, 3 : 4 : : 6 : 8. The symbol (: :) means “as” or “equals.” Either this symbol or the equal sign (=) ma y be used . The “extremes’ are th e first and las t terms. The “means” are the second and third terms.
(2) Rule 1. In proportion, the product of the means equals the product of the extremes.
EXAMPLE
Therefore, 3 : 4 : : 9 : 12.
4 x 9 = 36 Multiply the means.
3 x 12 = 36 Multiply the extremes. Thus, the two expressions are equal.
NOTE : This makes it possible to find an unknown quantity. In other words, when three terms of a proportion are known the fourth can be found.
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(3) Rule 2. To find one unknown mean. When both extremes and one mean
300 = 5X and means. Divide both sides 60 = X of the equation by 5. The
unknown mean is 60.
(4) Rule 3. To find one unknown extreme. When both means and one extreme are known, find the unknown extreme by dividing the product of the means by the known extreme. when a ratio is equal to its inverse, the elements are said to be inversely proportional.
(2) Two numbers are inversely proportional when one increases as the other decreases. In this case their product is always the same. A practical example of inverse ratio is seen in problems dealing with pulleys.
(a) Rule 1. The speed of pulleys connected by belts are inversely proportional to their diameters. The smaller pulley rotates faster then the larger pulley.
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EXAMPLE
A 24 inch pulley fixe d to a live shaft which make s 400 revo lutions per minute (rpm) is belted to a 6 inch pulley, as shown in figure 8 on the following page. Find the rpm of the smaller pulley.
This is what the problem looks like:
A is the driving pulley, B is the driven pulley.
Then, X : 400 :: 24 : 6
(b) Rule 2. The speed of gears ru nning together is inversely proportional to their number of teeth.
EXAMPLE
A driving gear with 48 teeth meshes with a driven gear which has 16 teeth.
If the driving gear makes 100 rpm, find the number of rpm of the driven gear.
3. Pulley Trains and Gear Trains
a. In the pre vious paragraph, we discussed the mea nings and met hods of solving ratio and proportion problems. In this paragraph, we will apply these methods to help determine the size to which a pulley or gear should be machined in order to enable it to rotate at a given number of revolutions per minute (rpm) for efficient operation of the machinery pulley train or vehicle gear tra in. A pulley train is a serie s of pull eys connected by belting as shown in figure 9 on page 47. A gear train is a series of gears running
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FIGURE 8. PULLEYS.
together; the power com es from on e of the pulle ys or gear s. Neglecting slippage of the belting in a pulley train, the same method of determining relative sizes applies to both systems.
EXAMPLE
Find the rpm of the 6 inch pulley shown in figure 9 on the following page.
Therefore
b. Screw Gearing .
(1) Spiral. Gears are often used to reduce speed. The teeth on the gears are arranged in the same manner as the threads of a screw. A spiral gear may have any number of teeth. A one-toothed gear corresponds to a single-threaded screw. A
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FIGURE 9. PULLEY TRAIN.
many-toothed gear corresponds to a many-threaded screw. Look at figure 10.
Count the teeth in the upper gear. There are 12. This gear, then, equals a 12-threaded screw. The lower gear has 36 teeth. It corresponds to a 36-threaded screw. Hence, the small gear mak es 3 complete turns while the large gear is making 1.
FIGURE 10. SPIRAL GEAR SYSTEM.
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(2) Worm and Worm Gearing . Worm gearing is used to transmit power between two shafts at 90° to each other, but not in th e same plane. In worm gearing, the velocity ratio is the ratio between the number of teeth on the gears and the number of threads on the worm. Figure 11 shows a worm and a single-threaded worm gear.
EXAMPLE
Find the revolutions per minute (rpm) for the worm gear.
20 x 60 = 1200 Convert 20 revolutions-per-second of the single worm to rpm by multiplying by 60 seconds.
Therefore
FIGURE 11. WORM GEAR SYSTEM.
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LESSON 2 CONVERSION OF LINEAR
MEASUREMENTS FROM THE ENGLISH TO THE METRIC SYSTEM AND FROM THE METRIC TO THE ENGLISH SYSTEM; AND SOLVING PROBLEMS USING
RATIO, PROPORTION, AND TRIGONOMETRY
TASK 3. Describe the processes for solving problems using trigonometry.
CONDITIONS
Within a self-study environment and given the subcourse text, without assistance.
STANDARDS Within two hours
REFERENCES
No supplementary references are needed for this task.
1. Introduction
In this task we will discuss the processes involved in solving machine shop problems through the use of trigonometry.
2. General
Task 1 and 2 of this lesson served respectively to describe the processes for converting linear measurements from the English to the metric system and from the metric to the English system, and for solving problems using ratio and proportion. In view of the fact that some military equipment has been developed with both metric and English measured components, Task 1 enables the machinist to convert linear measurements from one system to another, thereby ensuring the proper mating of machined parts that must operate in mesh with each other. Task 2 enables the machinist to determine the size that a spool, pulley or gear must be machined to, to permit its rotation at a specified
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rpm in the pulley or gear train of military machinery and vehicles.
Task 3 will describe the processes for solving problems through triangulation, otherwise known as trigonometry. Trigonometry is essentially that branch of mathematics which deals with the relations existing between the sides and angles of triangles. In this task only right triangles will be discussed. A right triangle is a triangle that contains one 90° angle and two other lesser angles for a total of 180° or half the number of degrees in a circle, which contains 360°.
This process will assist the machinist in determining the pitch or angle of screw threads, gear teeth, and tapers for parts that must be fabricated for items not normally available through supply channels, or in an emergency in a combat sit uation. Before goin g into the solv ing of tri gonometric problems, let’s first review the trigonometric functions which are the basis for solving these types of problems.
3. Trigonometric F unctions
a. For any giv en ac ute angle in a ri ght tri angle, ce rtain rat ios exist among the sides. These ratios are called “trigonometric functions.” They determine sides and angles in a right triangle. To this end, the sides of a right triangle are given certain names to indicate their relation to the angles. Thus, in any ri ght triangle, such as sh own in fi gure 12 on the following page, the side “c,” which is opposite to the right angle “C,” is called the “hypotenuse”; side “a” is opposite angle “A” and is called the
“opposite side”; side “b,” is adjacent to angle “A” and is called the
“adjacent side.” Notice, however, that when the sides refer to angle “B,”
side “b” is the opposite side and side “a” is the adjacent side. However, the hypotenuse, the longest side, is always called the hypotenuse with reference to either angle.
b. In this triangle, it is po ssible to show six different ratios of th e sides. They are a/c, b/c, a/b, b/a, c/b, and c/a. An explanation of these ratios follows using the ratio a/c as an example. This explanation is also applicable to the other ratios; a/c means the same as “a” divided by “c,”
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FIGURE 12. SIDES IN REFERENCE TO ANGLE “A.”
or “a” over “c.” In each ratio the letter in the same position as “a” in relation to “c” represents the numerator, which can be given a numerical value. The letter “c” in relation to the position of “a” represents the denominator, which can be given a numerical value. If the letter “a” is assigned the numerical value of 2 and the letter “c” is assigned the numerical value of 4, the mathematical expression would be that the number 2 must be divided by the number 4. Thus, 2/4 = .5. These ratios are the trigonometric functions as described below:
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c. The trigonometric functions discussed here will be li mited to the sine, cosine, tangent, and cotangent, since practically every common shop problem in trigonometry can be sol ved by mea ns of these fu nctions. The values of the trigonometric functions in terms of the names of the sides should be learned. To assist in learning these functions, use the example below.
EXAMPLE
Using the lettered and named sides of the triangle (figure 12 on the previous page), write the ratios for sin A, cos A, tan A, cot A, sin B, cos B, tan B, and cot B.
sin A = a/c; cos A = b/c; sin B = b/c;
cos B = a/c; and tan A = a/b; cot A = b/a;
tan B = b/a; cot B = a/b.
Therefore sin A = cos B and cos A = sin B tan A = cot B and cot A = tan B
d. A tri gonometric fu nction ex presses the v alue of an an gle in ter ms of the sides of the right triangle containing that angle. For instance, the value of angle A in figure 13 on the following page may be expressed as:
Thus, if the function and the dimension of one of the sides of that function ratio are known, then the dimension of the other side can be found.
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FIGURE 13. VALUE OF AN ANGLE IN TERMS OF THE SIDES.
e. The rules for identifying angles and sides of right triangles are:
(1)
(2)
(3)
(4)
(5) Side opposite = hypotenuse x sine (6) Side opposite = side adjacent x tangent (7) Side opposite = side adjacent ÷ cotangent
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(10) Side adjacent = side opposite ÷ tangent
(11) Hypotenuse = side opposite ÷ sine
(12) Hypotenuse = side adjacent ÷ cosine f. Procedure for using these Rules.
(1) In a right triangle, both the known and unknown sides (opposite, adjacent, and hypotenuse) of the problem are named.
(2) Choose from among the prev ious rules; select one that fits the give n numerical values.
(3) Substitute the given values in the rule and solve for the unknown.
EXAMPLE
Find side “a” if sin A = 3/5 and side “c” = 20.5 (figure 14 on the following page). Here the sine of angle A is given, and “a” is the side opposite.
According to rule (5) in paragraph 3e on page 53, side opposite = hypotenuse x sine. Substituting 20.5 for hypotenuse and 3/5 for sine, we get: side opposite = 20.5 x 3/5 = 12.3.
According to rule (5) in paragraph 3e on page 53, side opposite = hypotenuse x sine. Substituting 20.5 for hypotenuse and 3/5 for sine, we get: side opposite = 20.5 x 3/5 = 12.3.